Algebra I : Semester II : Polynomials

Sections:

Introduction  |   Section 1  |   Section 2  |   Section 3  |  Section 4  |   Section 5  |  Section 6

Section Six

Part 1  |  Part 2

Algebra 1: Section 6: Polynomials Factoring Perfect Square Trinomials Suppose that the area of the square above is x2 + 6x + 9. What would be the binomial representation of the length of a side? You can answer this question by looking back at the rules for special products.   Special Products Square of a Sum             (a + b)2 = a2 + 2ab + b2 Square of a Difference      (a – b)2 = a2 – 2ab + b2 You know to find the area of a square by squaring the length of a side. So, to find the side, find out the binomial that would square to x2 + 6x + 9. This kind of polynomial is a perfect square trinomial. Each end is a square and the middle term is two times the square root of each end.   To Factor a2 + 2ab + b2 or a2 – 2ab + b2 Find the square root of a2. Find the square root of b2. Check to make sure that 2ab is the middle term. Write them in the parenthesis as (a + b)2 or (a – b)2. If in doubt, check by using FOIL. Example 1: Factoring a Square Trinomial Find the side of the square above by factoring x2 + 6x + 9. The square root of x2 is x The square root of 9 is3 2ab = 2(x)(3) = 6x, which is the middle term. Therefore, x2 + 6x + 9 = (x + 3)2 Each side of the square measures x + 3 Example 2: Factoring Perfect Square Trinomials Factor each of the following. x2 – 14x + 49 The square root of x2 is x. The square root of 49 is 72ab = 2(x)(7) = 14x, which is the middle term x2 – 14x + 49 = (x – 7)2 25d2+ 70d + 49 The square root of25d2 is 5d The square root of 49 is 72ab = 2(5d)(7) = 70d, which is the middle term 25d2+ 70d + 49 = (5d + 7)2 a2 + 12a + 25 The square root of  a2is a The square root of 25 is 2ab = 2(a)(5) = 10a, which is not the middle term This is not a perfect square trinomial and cannot be factored using this methodIf you check, it can’t be factored at all and is prime Example 3: Solving by Factoring Solve x2 + 25 = 10x by factoring. Subtract 10x:      x2 – 10x + 25 = 0 Factor:                (x – 5)2 = 0 Set = 0                x – 5 = 0   or   x – 5 = 0                             x = 5               x = 5                           Check:                 x2 + 25 = 10x                            (5)2 + 25 = 10(5)                             25 + 25 = 50                             50 = 50  √ Solution:             {5}                            In some texts, the solution will be called a double root of 5.   Practice Click to get a new problem. Factor the polynomial and then click to see the answer. Remember that your answer is still correct if you have the same binomials reversed. Solve each of the equations by factoring. You may have to use the GCF first. Equation Factored Solutions n2 – 2n + 1 = 0 (n – 1)(n – 1) = 0 (n - 1)2 = 0 {1} b2 + 81 = -18b b2 + 18b + 81 = 0 (b + 9)(b + 9) = 0 (b + 9)2 = 0 {-9} a3 = 10a2 – 25a a3 – 10a2 + 25a = 0 a(a2 – 10a + 25) = 0 a(a – 5)2 = 0 {0, 5} 2a2 – 24a + 72 = 0 2(a2 – 12a + 36) = 0 2(a – 6)2 = 0 {6}   Section 6 Homework (10 points) Find the section homework link to submit it for a grade. You may take this only one time, so check your understanding of the material before completing your homework, and do your best. Congratulations! You have completed Unit 7. Study for and take the Unit 7 Exam (100 points). Find the Module Exam link to submit it for a grade.

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